Q77 / 180 · COMEDK 2015 · Mathematics

$11^{3}+12^{3}+13^{3}+\ldots . .+20^{3}$ is

  • A
    an even integer.
  • Correct answer
    B
    an odd integer divisible by 5 .
  • C
    multiple of 10 .
  • D
    an odd integer but not a multiple of 5 .

Solution

We have, $$ \begin{aligned} 11^{3}+12^{3}+13^{3}+\ldots+20^{3} \\ =\left(1^{3}+2^{3}+\ldots+10^{3}+11^{3}+\right.&\left.12^{3}+\ldots .+20^{3}\right) \\ &-\left(1^{3}+2^{3}+\ldots .+10^{3}\right) \end{aligned} $$ $$ \begin{aligned} &=\Sigma(20)^{3}-\Sigma(10)^{3} \\ &=\left(\frac{20(20+1)}{2}\right)^{2}-\left(\frac{10(10+1)}{2}\right)^{2} \\ &=(10 \times 21)^{2}-(5 \times 11)^{2}=(210-55)(210+55) \\ &=155 \times 265=41075=5 \times 8215 \end{aligned} $$

Hence, the correct answer is option: 'B'.

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